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A0.160-kg hockey puck is moving on an icy, frictionless, horizontal sufface. At $t=0$ , the puck is moving to the right at 3.00 $\mathrm{m} / \mathrm{s}$ . (a) Calculate the velocity of the puck (magnitude and direction) after a force of 25.0 $\mathrm{N}$ directed to the right has been applied for 0.050 $\mathrm{s} .(\mathrm{b}) \mathrm{If}$ , instead, a force of 12.0 $\mathrm{N}$ directed to the left is applied from $t=0$ to $t=0.050 \mathrm{s}$ , what is the final velocity of the puck?

(a) $\overrightarrow{\mathrm{v}}_{f}=10.81 \mathrm{m} / \mathrm{s}$ to the right.

(b) $\overrightarrow{\mathrm{v}}_{f}=-0.75 \mathrm{m} / \mathrm{s}$ to the left

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Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Hope College

University of Sheffield

in this question. A hockey puck is moving to the right with the velocity off 3 m per second and it has a mass off 0.160 kg in the first item off this question. Ah, force off 25 mutants is apply it to the right on that puck and that force less for 0.5 seconds. What is the velocity of the park after that force is? Apply it well. To begin solving this item, we have to calculate what is the initial momentum off the park? Let me call the initial momentum. P zero and P zero is given by the mass off the park times its initial velocity just for clarity. To solve this question, I'm using the following reference frame. Everything that points to the right will be positive. And this is my reference frame to solve this question. So the initial momentum off the park will be given by its mass 0.160 times the velocity, which is three in my reference frame. Because the velocity points to the same direction as the positive direction off my reference frame. Then the initial momentum is 0.480 kg meters per second. Now we have to calculate what is the change, the variation in the momentum due to the application off that force. We can use this equation for that. So the variation in the momentum Delta P is given by the force which is 25 Newtons times the time that this forces apply it which is 0.5 seconds. And this results in a valuation in the momentum off 1.25 kg meters per second. Just an important detail. The force entered here with a positive sign because the force is pointing to the right which is the positive direction off my reference frame. Okay, the reform after applying that force, the final momentum let me call this PF is given by the initial momentum plus the variation in the momentum show. The final momentum is given by zero point 480 plus 1.25 on this results in the final momentum off 1.730 kg meters per second. Now to determine what is the final velocity, we can use the fact that the momentum is given by the mass times the velocity. So we say that the final momentum 1.730 is given by the mass off the puck times the final velocity. The mass of the park is well known. It's 0.160. So here we have 0.160 times the final velocity. The reform. The final velocity of the park is given by one 0.730 divided by 0.160 on. This is approximately 10 0.8 m per second. Notice that the velocity is positive and positive vectors in our reference frame means that the velocity is pointing to the right because our reference frame is positive when things point to the right and this is the answer to the first item off this question in the second night, and we have to perform a very similar calculation, but with a different force, which now is 12 new turns to the left. So we already know what is the initial momentum. Then we begin by calculating what is the variation in the momentum Delta P is given by the force, which now is 12 Newton's. But it's 12 Newtons of the left. So in our reference frame it means that we should plug in the force with a negative sign since it's pointing in the negative direction off our reference frame and then we multiply that forced by the interval of time which is 0.5 on these results in minus 0.60 kg meters per second. So now this is a variation of the momentum. Then the final momentum PF is given by the initial momentum plus the variation in the momentum on this is 0.48 minus 0.60 on these results minus 0.12 kg meters per second. So this is the final momentum after that force is applied. Now we use the fact that the momentum is given by the mass times the velocity to calculate what is the final velocity so minus 0.12 is equal to the mass off the puck, which is 0.160 times its final velocity v f. The reform, The final velocity off the puck is given by minus 0.12 divided by 0.160 on this results in the final velocity off minus 0.75 m per second. So now the final velocity is 0.75 m per second to the left. Since the velocity is negative and we are in a reference frame where negative things points the left on. This is the answer to this question.

Brazilian Center for Research in Physics